Motivation.

A problem from this years phy1530 problem set 2 that appears appropriate for phy454 exam prep.

Statement.

Consider the steady flow between two long cylinders of radii and , , rotating about their axes with angular velocities , . Look for a solution of the form, where is a unit vector along the azimuthal direction:

\begin{subequations}

\end{subequations}

Write out the Navier-Stokes equations and find differential equations for and . You should find that these equations have relatively simple solutions, i.e.,

Fix gthe constants and from the boundary conditions. Determine the pressure .

Compute the friction forces that the fluid exerts on the cylinders, and compute the torque on each cylinder. Show that the total torque on the fluid is zero (as must be the case).

This is also a problem that I recall was outlined in section 2 from [1]. Some of the instabilities that are mentioned in the text are nicely illustrated in [2].

We illustrate our system in figure (1).

Figure 1: Coutette flow configuration

Solution: Part 1. Navier-Stokes and resulting differential equations.

Navier-Stokes for steady state incompressible flow has the form

\begin{subequations}

\end{subequations}

where the gradient has the form

Let’s first verify that the incompressible condition 3.3b is satisfied for the presumed form of the solution we seek. We have

Good. Now let’s write out the terms of the momentum conservation equation 3.3a. We’ve got

and

and

Equating and components we have two equations to solve

\begin{subequations}

\end{subequations}

Expanding out our velocity equation we have

for which we’ve been told to expect that 2.2 is a solution (and it has the two integration constants we require for a solution to a homogeneous equation of this form). Let’s verify that we’ve computed the correct differential equation for the problem by trying this solution

Given the velocity, we can now determine the pressure up to a constant

so

Solution: Part 2. Fixing the constants.

To determine our integration constants we recall that velocity associated with a radial position in cylindrical coordinates takes the form

where is the angular velocity. The cylinder walls therefore have the velocity

so our boundary conditions (given a no-slip assumption for the fluids) are

This gives us a pair of equations to solve for and

Multipling each by and respectively gives us

Rearranging for we find

or

For we have

or

This gives us

\begin{subequations}

\end{subequations}

Solution: Part 3. Friction and torque.

We can expand out the identity for the traction vector

in cylindrical coordinates and find

\begin{subequations}

\end{subequations}

so we have

\begin{subequations}

\end{subequations}

We want to expand the last of these

So the traction vector, our force per unit area on the fluid at the inner surface (where the normal is ), is

and our torque per unit area from the inner cylinder on the fluid is thus

Observing that our stress tensors flip sign for an inwards normal, our torque per unit area from the outer cylinder is

For the complete torque on the fluid due to a strip of width the magnitudes of the total torque from each cylinder are respectively

As expected these torques on the fluids sum to zero

Evaluating these at and respectively gives us the torques on the fluid by the cylinders, so inverting these provides the torques on the cylinders by the fluid

Motivation.

Exersize 6.1 from [1] is to show that the traction vector can be written in vector form (a rather curious thing to have to say) as

Note that the text uses a wedge symbol for the cross product, and I’ve switched to standard notation. I’ve done so because the use of a Geometric-Algebra wedge product also can be used to express this relationship, in which case we would write

In either case we have

(where the primes indicate the scope of the gradient, showing here that we are operating only on , and not ).

After computing this, lets also compute the stress tensor in cylindrical and spherical coordinates (a portion of that is also problem 6.10), something that this allows us to do fairly easily without having to deal with the second order terms that we encountered doing this by computing the difference of squared displacements.

We’ll work primarily with just the strain tensor portion of the traction vector expressions above, calculating

We’ll see that this gives us a nice way to interpret these tensor relationships. The interpretation was less clear when we computed this from the second order difference method, but here we see that we are just looking at the components of the force in each of the respective directions, dependent on which way our normal is specified.

Verifying the relationship.

Let’s start with the the plain old cross product version

We can also put the double cross product in wedge product form

Equivalently (and easier) we can just expand the dot product of the wedge and the vector using the relationship

We can now move on to compute the directional derivatives and complete the strain calculation in cylindrical coordinates. Let’s consider this computation of the stress for normals in each direction in term.

With .

Our directional derivative component for a normal direction doesn’t have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

With .

Our directional derivative component for a normal direction will have some cross terms since both and are functions of

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

With .

Like the normal direction, our directional derivative component for a normal direction will not have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

Summary.

\begin{subequations}

\end{subequations}

Spherical strain tensor.

Having done a first order cylindrical derivation of the strain tensor, let’s also do the spherical case for completeness. Would this have much utility in fluids? Perhaps for flow over a spherical barrier?

We need the gradient in spherical coordinates. Recall that our spherical coordinate velocity was

and our gradient mirrors this structure

We also previously calculated \inbookref{phy454:continuumL2}{eqn:continuumL2:1010} the unit vector differentials

\begin{subequations}

\end{subequations}

and can use those to read off the partials of all the unit vectors

Finally, our velocity in spherical coordinates is just

from which we can now compute the curl, and the directional derivative. Starting with the curl we have

So we have

With .

The directional derivative portion of our strain is

The other portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

This is consistent with (15.20) from [3] (after adjusting for minor notational differences).

With .

Now let’s do the direction. The directional derivative portion of our strain will be a bit more work to compute because we have variation of the unit vectors

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

This again is consistent with (15.20) from [3].

With .

Finally, let’s do the direction. This directional derivative portion of our strain will also be a bit more work to compute because we have variation of the unit vectors

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Disclaimer.

Review. Rayleigh Benard Problem

Reading: section 9.3 from [1].

Illustrated in figure (1) is the heated channel we’ve been discussing

Channel with heat applied to the base

We’ll take initial conditions

Our energy equation is

We have this term because our heat can be carried from one place to the other, due to the fluid motion. We’d not have this convective term for heat dissipation in solids because elements of a solid are not moving around in the bulk.

We’ll also use

In the steady (base) state we have

but since we are only considering spatial variation with we have

with solution

We found that after application of the perturbation

to the base state equations, our perturbed Navier-Stokes equation was

Application of the perturbation to the energy equation.

We’ve got

Using this, and 2.7, and neglecting any terms of second order smallness we have

but performing this non-dimensionalization shows that this was either quoted incorrectly, or typed wrong in the heat of the moment. A check against the text shows (equation (9.27)), shows that 3.23 is correct.

Non-dimensionalization of the energy equation

Rescaling our energy equation 3.16 we find

Introducing the \textit{Prandtl number}

and dropping primes our non-dimensionalized energy equation takes the form

Normal mode analysis.

We’ve got a pair of nasty looking coupled equations 3.24, and 3.27. Repeated so that we can see them together

\begin{subequations}

\end{subequations}

it’s clear that we can decouple these by inserting 3.42 into 3.28a. Doing that gives us a beastly 6th order spatial equation for the perturbed temperature

It’s pointed out in the text we have all the and derivatives coming together we can apply separation of variables with

provided we introduce some restrictions on the form of . Here (if real) is the growth rate. Applying the Laplacian to this assumed solution we find

where

For 3.31 to be separable we require a constant proportionality

or

Picking so that we don’t have hyperbolic solutions, must have the form

where

Our separation of variables function now takes the form

Writing

our beastly equation to solve is then given by

This is now an equation for only

Conceptually we have just a plain old LDE, and should we decide to expand this out we have something of the form

Our standard toolbox method to solve this is to assume a solution and compute the characteristic equation. We’d have to solve

Let’s back up a bit instead. Looking back to 3.42, it’s clear that we’ll have the same separable form for our perturbed velocity since we have

where

Assuming a solution of the form

our velocity is then fully specified in terms of the temperature, since we have

Back to our coupled equations.

Having gleamed an idea what the form of our solutions is, we can simplify our original coupled system, writing

\begin{subequations}

\end{subequations}

Considering the boundary conditions, if the heating is even then at we can’t have any variation with and , so can only have . Thus at the boundary, from 3.46a, we have

From the continuity equation , the text argues that we also have on the boundary, so that on that plane we also have

Expanding out 3.47 then gives us

or

These boundary value constraints 3.48, and 3.50, plus the coupled system equations 3.46 are the complete problem to solve. To get a feel for the solution of this system, consider the system with the following simpler set of boundary value constraints

which in the text is described as the artificial problem of thermal instability for boundaries that are stress free (FIXME: it’s not clear to me what that means without some thought … return to this). For such a system on the boundaries (noting that we are still in dimensionless quantities), we have solutions

Note that we have

Inserting 3.52 into our system 3.46, we have

For any , , we must then have

For , this gives us the critical value for the Rayleigh number

the value that separates our stable and unstable solutions. On the other hand for

(), we have an instable system.

(), we have a stable system.

FIXME: The text has a positive sign on the term above. He actually solves the quadratic for , but I don’t see how that would make a difference. Is there an error here, or a typo in the text?

This is illustrated in figure (2).

Critical Rayleigh's number.

The instability means that we’ll have instable flows as illustrated in figure (3)

Instability due to heating.

Solving for these critical points we find

\begin{subequations}

\end{subequations}

Multimedia presentations.

Kelvin-Helmholtz instability.Colored salt water underneath, with unsalty water on top. Apparatus tilted causing flow of one over the other. Instability of the interface.

See [2] for a really cool animation of a simulation of this effect. It ends up looking very fractal. Also interesting is the picture of this observed for real in the atmosphere of Saturn.

A simulated mushroom cloud occurring with one fluid seeping into another. This looks it matches what we find under Rayleigh-Taylor instability in [3].

plume, motion up through a denser fluid.

Plateau-Rayleigh instability. Drop pinching off. See instability in the fluid channel feeding the drop. A crude illustration of this can be found in figure (4).

Crude illustration of instability leading to a drop pinching off.

A better illustrations (and animations) can be found in [4].

Jet of water injected into a rotating tub on a turntable. Jet forms and surfaces.

we were able to show that inviscid irrotational incompressible flows are governed by Bernoulli’s equation

In the absence of body forces (or constant potentials), we have

so that our boundary layer equations are

\begin{subequations}

\end{subequations}

With boundary conditions

Fluid flow over a flat plate (Blasius problem).

Define a similarity variable

Suppose we want

Since we have

or

We can make the transformation

We can introduce stream functions

We can check that this satisfies the continuity equation since we have

Now introduce a similarity variable

Note that we’ve also suddenly assumed that (a constant, which will also kill the term in the N-S equation). This isn’t really justified by anything we’ve done so far, but asking about this in class, it was stated that this is a restriction for this formulation of the Blasius problem.

Also note that this last step requires:

This at least makes sense dimensionally since we have , but where did this definition of come from?

In [] it is mentioned that this is a result of the scaling argument. We did have some scaling arguments that included in the expressions from last lecture, one of which was 2.7

but that doesn’t obviously give us 3.20?

Ah. We argued that

and that the larger of the viscous terms was

If we require that these are the same order of magnitude, as argued in section 8.3 of [2], then we find 3.21.

Deriving the equation of motion.

Attempting to derive 3.24 using the definitions above gets a bit messy. It’s messy enough that I mistakenly thought that we couldn’t possible arrive at a differential equation that has a plain old (non-derivative) in it as in 3.24 above. The algebra involved in taking the derivatives got the better of me. This derivation is treated a different way in [2]. For the purpose of completeness (and because that derivation also leaves out some details), lets do this from start to finish with all the gory details, but following the outline provided in the text.

Instead of pre-determining the form of the similarity variable exactly, we can state it in terms of an unknown and to be determined function of position writing

We still introduce stream our stream functions 3.17, but require that our horizontal velocity component is only a function of our similarity variable

where is to be determined, and is scaled by our characteristic velocity . Observe that, as above, we are assuming that , a constant (which also kills off the term in the Navier-Stokes equation.) Given this form of , we note that

so that

It’s argued in the text that we also want to be a streamline, so that implying that . I don’t honestly follow the rational for that, but it’s certainly convenient to set , so lets do so and see where things go. With

Observe that is necessarily zero with this definition. We can now write

This is like what we had in class, with the exception that instead of a constant relating and we also have a function of . That’s exactly what we need so that we can end up with both and derivatives of in our Navier-Stokes equation.

Now let’s do the mechanical bits, computing all the derivatives. We can compute to start with

We had initially , but , so we’ve now got both and specified in terms of and and their derivatives

We’ve got a bunch of the derivatives that we have to compute

and

and

Our component of Navier-Stokes now takes the form

or (assuming )

Now, it we wish (to make this equation as easy to solve as possible), we can integrate to find the required form of . This gives

It’s argued that we expect

to become singular at , so we should set . This leaves us with

\begin{subequations}

\end{subequations}

and boundary value conditions

\begin{subequations}

\end{subequations}

where 3.38b follows from and 3.37d, and 3.38c follows from the fact that tends to .

Numeric solution.

We can solve this numerically and find solutions that look like figure (2).

Figure 2: Boundary layer solution to flow over plate.

This is the Blasius solution to the problem of fluid flow over a flat plate.

Singular perturbation theory.

In the boundary layer analysis we’ve assumed that our inertial term and viscous terms were of the same order of magnitude. Lets examine the validity of this assumption

or

or

finally

If we have

and

then

when .

(this is the whole reason that we were able to do the previous analysis).

Our EOM is

with

as

performing a non-dimensionalization we have

or

to force , we can write

so that as we have .

With a very small number modifying the highest degree partial term, we have a class of differential equations that doesn’t end up converging should we attempt a standard perturbation treatment. An example that is analogous is the differential equation

where and . The exact solution of this ill conditioned differential equation is

This is illustrated in figure (3).

Figure 3: Solution to the ill conditioned first order differential equation.

Study of this class of problems is called \textit{Singular perturbation theory}.

Motivation and statement.

Viscous fluid is at rest in a two-dimensional channel between stationary rigid walls with . For a constant pressure gradient is imposed. Show that satifies

and give suitable initial and boundary conditions. Find in form of a Fourier series, and show that the flow approximates to steady channel flow when .

Solution

With only horizontal components to the flow, the Navier-Stokes equations for incompressible flow are

\begin{subequations}

\end{subequations}

Substitution of 2.2b into 2.2a gives us

Our equation to solve is therefore

This equation, for , allows for solutions

but the problem states that the fluid is at rest initially, so we don’t really have to solve anything (i.e. ).

The no-slip conditions introduce boundary value conditions .

For we have

If we attempt separation of variables with , our equation takes the form

We see that the non-homogenous term prevents successful application of separation of variables. Let’s modify our problem by attempting to recast our equation into a homogenous form by adding a particular solution for the steady state flow problem. That problem was the solution of

which has solution

The freedom to incoporate an constant into the equation as an integration constant has been employed, knowing that it will kill the contributions at to make the boundary condition matching easier. Our no-slip conditions give us

Adding this we have , and subtracting gives us , so a specific solution that matches our required boundary value (and initial value) conditions is just the steady state channel flow solution we are familiar with

Let’s now assume that our general solution has the form

Applying the Navier-Stokes equation to this gives us

But from 2.8, we see that all we have left is a homogenous problem in

where our boundary value conditions are now given by

and

or
\begin{subequations}

\end{subequations}

Now we can apply separation of variables with , yielding

or

Here a positive constant has been used assuming that we want a solution that is damped with time.

Our solutions are

or

We have constraints on due to our boundary value conditions. For our sin terms to be solutions we require

Rather remarkably, this Fourier series is actually a very good fit even after only a single term. Using the viscosity and density of water, , and (parameterizing the pressure gradient by the average velocity it will induce), a plot of the parabola that we are fitting to and the difference of that from the first Fourier term is shown in figure (1).

The higher order corrections are even smaller. Even the first order deviations from the parabola that we are fitting to is a correction on the scale of of the height of the parabola. This is illustrated in figure (2) where the magnitude of the first 5 deviations from the steady state are plotted.

Figure 2: Difference from the steady state for the first five Fourier terms.

An animation of the time evolution above can be found in figure (3). If this animation is unavailable, it can also be found at http://youtu.be/0vZuv9HBtmo.

Figure 3: Time evolution of channel flow velocity profile after turning on a constant pressure gradient.

It’s also interesting to look at the very earliest part of the time evolution. Observe in figure (4) (or http://youtu.be/dDkx8iLwOew) the oscillatory phenomina. Could some of that be due to not running with enough Fourier terms in this early part of the evolution when more terms are probably significant?

Figure 4: Early time evolution of channel flow velocity profile after turning on a constant pressure gradient.

Disclaimer.

Two ill conditioned LDEs.

With , and letting , we’ll look at solutions of the ill conditioned LDE

With , , and we’ll look at the second order ill conditioned LDE

The first order LDE.

Exact solution.

Our homogeneous equation is

with solution

Looking for a solution of the form

we find

and integrate to find

Application of the boundary value constraints give us

This is plotted in figure (1).

Figure 1: Plot of exact solution to simple first order ill conditioned LDE.

Limiting cases.

We want to consider the limiting case where

and we let . If , then we have

or just

However, if then we have to be more careful constructing an approximation. When is very small, but is also of the same order of smallness we have

If and

so

Approximate solution in the inner region.

When define a new scale

so that our LDE takes the form

When we have

We have solution

or

Question: Couldn’t we just Laplace transform.Answer given: We’d still get into trouble when we take . My comment: I don’t think that’s strictly true. In an example like this where we have an exact solution, a Laplace transform technique should also yield that solution. I think the real trouble will come when we attempt to incorporate the non-linear inertial terms of the Navier-Stokes equation.

Second order example.

Exact solution.

We saw above in the first order system that our specific solution was polynomial. While that was found by the method of variation of parameters, it seems obvious in retrospect. Let’s start by looking for such a solution, starting with a first order polynomial

Application of our LDE operator on this produces

Now let’s move on to find a solution to the homogeneous equation

As usual, we look for the characteristic equation by assuming a solution of the form . This gives us

with roots

So our homogeneous equation has the form

and our full solution is

with the constants and to be determined from our boundary value conditions. We find

We’ve got and by subtracting

So the exact solution is

This is plotted in figure (2).

Figure 2: Plot of our ill conditioned second order LDE

Solution in the regular region.

For small relative to our LDE is approximately

which has solution

Our boundary value constraint gives us

Our solution in the regular region where and is therefore just

Solution in the ill conditioned region.

Now let’s consider the inner region. We’ll see below that when , and we allow both and tend to zero independently, we have approximately

We’ll now show this. We start with a helpful change of variables as we did in the first order case

When and we have

or

This puts the LDE into a non ill conditioned form, and allows us to let . We have approximately

We’ve solved this in our exact solution work above (in a slightly more general form), and thus in this case we have just

at we have

so that

and we find for the inner region

Taking these independent solutions for the inner and outer regions and putting them together into a coherent form (called matched asymptotic expansion) is a rich and tricky field. For info on that we’ve been referred to [1].

Disclaimer.

This problem set is as yet ungraded.

Problem Q1.

Background.

In fluid convection problems one can make several choices for characteristic velocity scales. Some choices are given below for example:

where is the acceleration due to gravity, is the coefficient of volume expansion, length scale associated with the problem, is the applied temperature difference, is the kinematic viscosity and is the thermal diffusivity.

Part 1. Statement. Check whether the dimensions match in each case above.

Solution.

Part 2. Statement. Pure liquid.

For pure liquid, say pure water at room temperature, one has the following estimates in cgs units:

For a layer depth and a ten degree temperature drop convective velocities have been experimentally measured of about .

With , calculate the values of , , , and . Which ones of the characteristic velocities , , do you think are suitable for nondimensionalising the velocity in Navier-Stokes/Energy equation describing the water convection problem?

We have

Use of gives the closest match to the measured characteristic velocity of .

Solution.

Part 3. Statement. Mantle convection.

For mantle convection, we have

and the actual convective mantle velocity is . Which of the characteristic velocities should we use to nondimensionalise Navier-Stokes/Energy equations describing mantle convection?

Solution

Let’s compute the characteristic velocities again with the mantle numbers

Both and come close to the actual convective mantle velocity of . Use of to nondimensionalise is probably best, since it has more degrees of freedom, and includes the gravity term that is probably important for such large masses.

Problem Q2.

Statement

Nondimensionalise N-S equation

where is the unit vector in the direction. You may scale:

velocity with the characteristic velocity ,

time with , where is the characteristic length scale,

pressure with ,

Reynolds number and Froude number .

Solution

Let’s start by dividing by , to make all terms (most obviously the term) dimensionless.

Our suggested replacements are

Plugging these in we have

Making a replacement, using the Froude number, we have

Scaling by we tidy things up a bit, and also allow for insertion of the Reynold’s number

Observe that the dimensions of Froude’s number is that of velocity

so that the end result is dimensionless as desired. We also see that Froude’s number, characterizes the significance of the body force for fluid flow at the characteristic velocity. This is consistent with [1] where it was stated that the Froude number is used to determine the resistance of a partially submerged object moving through water, and permits the comparison of objects of different sizes (complete with pictures of canoes of various sizes that Froude built for such study).

Problem Q3.

Statement

In case of Stokes’ boundary layer problem (see class note) calculate shear stress on the plate . What is the phase difference between the velocity of the plate and the shear stress on the plate?

Solution

We found in class that the velocity of the fluid was given by

where

Calculating our shear stress we find

and on the plate () this is just

We’ve got a constant term, plus one that is sinusoidal. Observing that

The phase difference between the non-constant portion of the shear stress at the plate, and the plate velocity is just . The shear stress at the plate lags the driving velocity by 90 degrees.

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Stability. Some graphical illustrations.

What do we mean by stability? A configuration is stable if after a small disturbance it returns to it’s original position. A couple systems to consider are shown in figures (\ref{fig:continuumL21:continuumL21Fig1a}), (\ref{fig:continuumL21:continuumL21Fig1b}) and (\ref{fig:continuumL21:continuumL21Fig1c}).

Figure 1: Stable well configuration.

Figure 2: Instable peak configuration.

Figure 3: Stable tabletop configuration.

We can examine how a displacement changes with time after making it. In a stable configuration without friction we will induce an oscillation as plotted in figure (\ref{fig:continuumL21:continuumL21Fig2a}) for the parabolic configuration. With friction we’ll have a damping effect. This is plotted for the parabolic well in figure (\ref{fig:continuumL21:continuumL21Fig2b}).

Figure 4: Displacement time evolution in undamped well system.

Figure 5: Displacement time evolution in damped well system.

For the inverted parabola our displacement takes the form of figure (\ref{fig:continuumL21:continuumL21Fig3})

Figure 6: Time evolution of displacement in instable parabolic configuration.

For the ball on the table, assuming some friction that stops the ball, fairly quickly, we’ll have a displacement as illustrated in figure (\ref{fig:continuumL21:continuumL21Fig3b})

Figure 7: Time evolution of displacement in tabletop configuration.

Characterizing stability

Let’s suppose that our displacement can be described in exponential form

where is the \textit{growth rate of perturbation}, and is in general a complex number of the form

Case I. Oscillatory unstability

A system of the form

\textit{oscillatory unstable}. An example of this is the undamped parabolic system illustrated above.

Case II. Marginal unstability.

We’ll call systems of the form

\textit{marginally unstable}. We can have unstable systems with but still , but these are less common.

Case III. Neutral stability.

An example of this was the billiard table example where the ball moved to a new location on the table after being bumped slightly.

A mathematical description.

For a discussion of stability in fluids we’ll not only have to incorporate the Navier-Stokes equation as we’ve done, but will also have to bring in the heat equation. Unfortunately that isn’t in the scope of this course to derive. Let’s consider as system heated on a bottom plate, and consider the fluid and convection due to heating. This system is illustrated in figure (\ref{fig:continuumL21:continuumL21Fig4})

Figure 8: Fluid in cavity heated on the bottom plate.

We start with Navier-Stokes as normal

For steady state with initially (our base state), we’ll call the following the equation of the base state

We’ll allow perturbations of each of our variables

After perturbation NS takes the form

Retaining only terms that are of first order of smallness.

applying our equation of base state 4.4, we have

or

we can write

Applying the divergence operation on both sides, and using so that we have

or

Assuming that is constant (actually that’s already been done above), we can cancel it, leaving

operating once more with we have

Going back to 4.9 and taking only the component we have

in the last step we use the following assumed relation for temperature

Here is the coefficient of thermal expansion. This is just a statement that expansion and temperature are related (as we heat something, it expands), with the ratio of the density change relative to the original being linearly related to the change in temperature.

We have finally

Solving this is the Rayleigh-Benard instability problem.

While this is a fourth order differential equation, it’s still the same sort of problem logically as we’ve been working on. Our boundary value conditions at are

Also relevant will be a similar equation relating temperature and fluid flow rate

Problem Q3 (revisited).

I’d produced the following sketches. For a higher viscosity bottom layer , this should look something like figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig4}) whereas for the higher viscosity on the top, these would be roughly flipped as in figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig5}).

This superposition can be justified since we have no term in the Navier-Stokes equations for these systems.

Exact solutions.

The figures above are kind of rough. It’s not actually hard to solve the system above. After some simplification, I find with Mathematica the following solution

Should we wish a more exact plot for any specific values of the viscosities, we could plot exactly with software the vector field described by these velocities.

I suppose it is cheating to use Mathematica and then say that the solution is easy? To make amends for being lazy with my algebra, let’s show that it is easy to do manually too. I’ll do the same problem manually, but generalize it slightly. We can do this easily if we just be a bit marter with our integration constants. Let’s solve the problem for the upper and lower walls moving with velocity and respectively, and let the heights from the interface be and respectively.

We have the same set of differential equations to solve, but now let’s write our solution with the undetermined coefficients expressed as

Now it’s super easy to match the boundary conditions at and (the lower and upper walls respectively). Clearly the integration constants are just the velocities. Matching the tangential component of the traction vectors at we have